Problem: $f(n)=41-5n$ Complete the recursive formula of $f(n)$. $f(1)=$
Answer: $f( 1)=41-5( 1)={36}$ $f( 2)=41-5( 2)={31}$ $f( 2)-f( 1)={31}-{36}={-5}$ So the first term of the sequence is ${36}$ and the common difference is ${-5}$. This is the recursive formula of the sequence: $\begin{cases} f(1)={36} \\\\ f(n)=f(n-1)+({-5}) \end{cases}$